What is the extraneous solution to these equations? $\dfrac{x^2 + 5x}{x + 1} = \dfrac{-6x - 10}{x + 1}$
Explanation: Multiply both sides by $x + 1$ $ \dfrac{x^2 + 5x}{x + 1} (x + 1) = \dfrac{-6x - 10}{x + 1} (x + 1)$ $ x^2 + 5x = -6x - 10$ Subtract $-6x - 10$ from both sides: $ x^2 + 5x - (-6x - 10) = -6x - 10 - (-6x - 10)$ $ x^2 + 5x + 6x + 10 = 0$ $ x^2 + 11x + 10 = 0$ Factor the expression: $ (x + 1)(x + 10) = 0$ Therefore $x = -1$ or $x = -10$ At $x = -1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -1$, it is an extraneous solution.